Three immiscible liquids of densities $d_1 > d_2 > d_3$ and refractive indices $\mu_1 > \mu_2 > \mu_3$ are put in a beaker. The height of each liquid column is $\frac{h}{3}$. $A$ dot is made at the bottom of the beaker. For near normal vision,find the apparent depth of the dot.

(N/A) The apparent depth of an object in a medium of refractive index $\mu_1$ when viewed from a medium of refractive index $\mu_2$ is given by $d' = d \times (\frac{\mu_2}{\mu_1})$.
Let the dot be at $P$. The observer is in the air $(\mu_{air} = 1)$.
$1$. Apparent depth of $P$ as seen from the second liquid $(\mu_2)$:
$x_1 = \frac{h}{3} \times (\frac{\mu_2}{\mu_1})$.
$2$. Apparent depth of the image $P_1$ as seen from the third liquid $(\mu_3)$:
The real depth is $h_2 = \frac{h}{3} + x_1$.
$x_2 = (\frac{h}{3} + x_1) \times (\frac{\mu_3}{\mu_2}) = \frac{h}{3} \times \frac{\mu_3}{\mu_2} + \frac{h}{3} \times \frac{\mu_3}{\mu_1}$.
$3$. Apparent depth of the image $P_2$ as seen from air $(\mu_{air} = 1)$:
The real depth is $h_3 = \frac{h}{3} + x_2$.
$x_3 = (\frac{h}{3} + x_2) \times (\frac{1}{\mu_3}) = \frac{h}{3} \times \frac{1}{\mu_3} + \frac{x_2}{\mu_3}$.
Substituting $x_2$:
$x_3 = \frac{h}{3} (\frac{1}{\mu_3} + \frac{1}{\mu_2} + \frac{1}{\mu_1})$.
Thus,the apparent depth is $\frac{h}{3} (\frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3})$.